Iterative Processes

## Iterative Processes

An iterative process is a process that repeats, using its own result value as the start value for the next repeat of the process. Quadratic and cubic equations can be rearranged to allow them to be solved using an iterative process.

Rearrange the equation to have the highest power of x on one side of the equation, and all other values on the other side. Square or cube root (as required) both sides of the equation such that one side of the equation has an x value only.

The starting value for each iteration is shown with a suffix as x_n: the result of each suffix is shown as x_(n+1).

With a starting value, evaluate the equation. This will be the starting value for the next iteration e.g.:

x^2 -3x - 10 = 0

Rearrange to x^2 = 3x + 10

Obtain x on one side:x = sqrt(3x + 10)

The iteration is: x_(n+1) = sqrt(3x_n + 10)

## Example 1

Starting with an initial value of 10, find the solution to x^2 -3x -10 = 0.

Rearrange to x = sqrt(3x + 10)

Iteration is x_(n+1) = sqrt(3x_n + 10)

First value is 10: x_1 = sqrt(3 xx 10 + 10) = 5.83095

Substitute new value into the iteration: x_2 = sqrt(3 x 5.83095 + 10) = 5.24336

Obtain x_3 = 5.07248 and repeat until x_7 = 5.00018

To three decimal places, the answer is 5.000.

Substitute into the original equation to check.

## Example 2

Find a positive solution to x^3 - 4x^2 - 11x +30 = 0. Use an initial value of 10.

Rearrange: x^3 = 4x^2 + 11x - 30

One side equal to x: x = ^3sqrt(4x^2 + 11x - 30)

Iteration is x_1 = ^3sqrt(4x_0^2 + 11x_0 - 30)

First value is 10:

x_1 = ^3sqrt(4 xx (10)^2 + 11 xx 10 - 30) = 7.82974

Substitute the new value:

x_2 = ^3sqrt(4 xx (7.82974)^2 + 11 xx 7.82974 - 30)

= 6.70433

and repeat until x_12 = 5.02828, x = 5.0 to 1dp